#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1 Circuit with multiple diodes: which lamps light up?
#@TaskText_HTML@#
The following simulation includes multiple diodes and several lamps.
A lamp lights brightly when a voltage of approximately
\[
\begin{align*}
U_{\rm lamp}\geq 5~{\rm V}
\end{align*}
\]
drops across it.
Close the switch in the simulation.
* Which lamps light up brightly?
* Which lamps remain dark?
* Explain the result using the idea of diode bypass paths.
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1. Determine which lamps light up brightly.
#@PathBegin_HTML~12001~@#
Number the lamps from left to right:
\[
\begin{align*}
L_1,\;L_2,\;L_3,\;L_4,\;L_5.
\end{align*}
\]
After closing the switch, check the voltage across each lamp in the simulation.
A lamp is assumed to light brightly if
\[
\begin{align*}
U_{\rm lamp}\geq 5~{\rm V}.
\end{align*}
\]
The simulation shows that the outer lamps have a sufficiently large voltage across them, while the inner lamps are bypassed by conducting diodes.
#@PathEnd_HTML@#
#@ResultBegin_HTML~12001~@#
The lamps
\[
\begin{align*}
L_1
\quad \text{and} \quad
L_5
\end{align*}
\]
light up brightly.
#@ResultEnd_HTML@#
2. Determine which lamps remain dark.
#@PathBegin_HTML~12002~@#
The inner lamps are connected in parts of the circuit that are bypassed by forward-biased diodes.
A forward-biased diode has only a small voltage drop. Therefore, a lamp in parallel with such a diode path receives only a small voltage.
If
\[
\begin{align*}
U_{\rm lamp}<5~{\rm V},
\end{align*}
\]
the lamp does not light brightly.
#@PathEnd_HTML@#
#@ResultBegin_HTML~12002~@#
The lamps
\[
\begin{align*}
L_2,\;L_3,\;L_4
\end{align*}
\]
remain dark or almost dark.
#@ResultEnd_HTML@#
#@TaskEnd_HTML@#
#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1 Circuit with multiple diodes II: current calculation
#@TaskText_HTML@#
The following simulation includes two diodes and two resistors.
Assume a simple constant-voltage diode model:
\[
\begin{align*}
U_{\rm F}=0.6~{\rm V}.
\end{align*}
\]
The source voltage is
\[
\begin{align*}
U_0=4.0~{\rm V}.
\end{align*}
\]
The resistors are
\[
\begin{align*}
R_1=200~\Omega,
\qquad
R_2=100~\Omega.
\end{align*}
\]
Calculate the currents through
* \(D_1\),
* \(R_1\),
* \(R_2\).
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1. Calculate the current through \(R_1\).
#@PathBegin_HTML~12004~@#
The current through \(R_1\) passes through one forward-biased diode.
Therefore the voltage across \(R_1\) is
\[
\begin{align*}
U_{R1}
=
U_0-U_{\rm F}.
\end{align*}
\]
Insert the values:
\[
\begin{align*}
U_{R1}
&=
4.0~{\rm V}-0.6~{\rm V}
\\
&=
3.4~{\rm V}.
\end{align*}
\]
Now apply Ohm's law:
\[
\begin{align*}
I_{R1}
=
\frac{U_{R1}}{R_1}
=
\frac{3.4~{\rm V}}{200~\Omega}
=
17~{\rm mA}.
\end{align*}
\]
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#@ResultBegin_HTML~12004~@#
\[
\begin{align*}
I_{R1}=17~{\rm mA}
\end{align*}
\]
#@ResultEnd_HTML@#
2. Calculate the current through \(R_2\).
#@PathBegin_HTML~12005~@#
The current through \(R_2\) passes through two forward-biased diodes.
Therefore the voltage across \(R_2\) is
\[
\begin{align*}
U_{R2}
=
U_0-2U_{\rm F}.
\end{align*}
\]
Insert the values:
\[
\begin{align*}
U_{R2}
&=
4.0~{\rm V}-2\cdot 0.6~{\rm V}
\\
&=
2.8~{\rm V}.
\end{align*}
\]
Now apply Ohm's law:
\[
\begin{align*}
I_{R2}
=
\frac{U_{R2}}{R_2}
=
\frac{2.8~{\rm V}}{100~\Omega}
=
28~{\rm mA}.
\end{align*}
\]
#@PathEnd_HTML@#
#@ResultBegin_HTML~12005~@#
\[
\begin{align*}
I_{R2}=28~{\rm mA}
\end{align*}
\]
#@ResultEnd_HTML@#
3. Calculate the current through \(D_1\).
#@PathBegin_HTML~12006~@#
The diode \(D_1\) supplies both current paths.
Therefore, by Kirchhoff's current law,
\[
\begin{align*}
I_{D1}
=
I_{R1}+I_{R2}.
\end{align*}
\]
Insert the values:
\[
\begin{align*}
I_{D1}
&=
17~{\rm mA}+28~{\rm mA}
\\
&=
45~{\rm mA}.
\end{align*}
\]
#@PathEnd_HTML@#
#@ResultBegin_HTML~12006~@#
\[
\begin{align*}
I_{D1}=45~{\rm mA}
\end{align*}
\]
#@ResultEnd_HTML@#
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#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1 Circuit with multiple diodes III: switch-dependent currents
#@TaskText_HTML@#
The following simulation includes two diodes and a switch.
Assume a simple constant-voltage diode model:
\[
\begin{align*}
U_{\rm F}=0.7~{\rm V}.
\end{align*}
\]
The source voltage is
\[
\begin{align*}
U_0=5.0~{\rm V}.
\end{align*}
\]
The resistor is
\[
\begin{align*}
R_1=1.0~{\rm k}\Omega.
\end{align*}
\]
Calculate the currents through
* \(R_1\),
* \(D_1\),
* \(D_2\),
depending on the switch state \(S\).
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1. Calculate the currents for open switch \(S\).
#@PathBegin_HTML~12007~@#
With the switch open, only \(D_1\) is connected to the resistor path.
The conducting diode clamps the node voltage to approximately
\[
\begin{align*}
U_{\rm node}\approx U_{\rm F}=0.7~{\rm V}.
\end{align*}
\]
The resistor current is therefore
\[
\begin{align*}
I_{R1}
&=
\frac{U_0-U_{\rm F}}{R_1}
\\
&=
\frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega}
\\
&=
4.3~{\rm mA}.
\end{align*}
\]
Since only \(D_1\) conducts,
\[
\begin{align*}
I_{D1}=I_{R1},
\qquad
I_{D2}=0.
\end{align*}
\]
#@PathEnd_HTML@#
#@ResultBegin_HTML~12007~@#
For open switch:
\[
\begin{align*}
I_{R1}&=4.3~{\rm mA},
\\
I_{D1}&=4.3~{\rm mA},
\\
I_{D2}&=0.
\end{align*}
\]
#@ResultEnd_HTML@#
2. Calculate the currents for closed switch \(S\).
#@PathBegin_HTML~12008~@#
With the switch closed, \(D_1\) and \(D_2\) are connected in parallel.
The resistor current is still determined by the source voltage, the forward diode voltage, and \(R_1\):
\[
\begin{align*}
I_{R1}
&=
\frac{U_0-U_{\rm F}}{R_1}
\\
&=
\frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega}
\\
&=
4.3~{\rm mA}.
\end{align*}
\]
Kirchhoff's current law gives
\[
\begin{align*}
I_{D1}+I_{D2}=I_{R1}.
\end{align*}
\]
With the ideal constant-voltage diode model, the individual currents through two parallel diodes are not uniquely determined.
If both real diodes are approximately identical, the current splits approximately equally:
\[
\begin{align*}
I_{D1}
\approx
I_{D2}
\approx
\frac{4.3~{\rm mA}}{2}
=
2.15~{\rm mA}.
\end{align*}
\]
#@PathEnd_HTML@#
#@ResultBegin_HTML~12008~@#
For closed switch:
\[
\begin{align*}
I_{R1}=4.3~{\rm mA}
\end{align*}
\]
and
\[
\begin{align*}
I_{D1}+I_{D2}=4.3~{\rm mA}.
\end{align*}
\]
For approximately identical real diodes:
\[
\begin{align*}
I_{D1}
\approx
I_{D2}
\approx
2.15~{\rm mA}.
\end{align*}
\]
#@ResultEnd_HTML@#
3. Explain why the current sharing is not unique in the simple model.
#@PathBegin_HTML~12009~@#
The constant-voltage diode model assumes that each conducting diode has exactly the same voltage drop:
\[
\begin{align*}
U_{\rm D}=U_{\rm F}.
\end{align*}
\]
For two parallel diodes, this condition is true for many possible current distributions.
Therefore, the model only determines the sum
\[
\begin{align*}
I_{D1}+I_{D2},
\end{align*}
\]
not the individual diode currents.
#@PathEnd_HTML@#
#@ResultBegin_HTML~12009~@#
The constant-voltage diode model determines only
\[
\begin{align*}
I_{D1}+I_{D2}=4.3~{\rm mA}.
\end{align*}
\]
It does not uniquely determine \(I_{D1}\) and \(I_{D2}\) separately.
Parallel diodes are sensitive to small differences in real diode characteristics.
Current sharing should not be assumed to be perfect without checking the design.
#@ResultEnd_HTML@#
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